看到个有意思的问题 “给定2个整数，这两个数大的连long类型都装不下，如何计算两数和”
实践了一下，可以用数组存放，数组的每个元素对应大整数的每一个数位
fun bigNumberSum(bigNumA:String, bigNumB:String){
val maxArrayLength =if (bigNumA.length > bigNumB.length) bigNumA.length +1 else bigNumB.length +1
val a = IntArray(maxArrayLength)
//倒序放入数组
for (iin 0 until bigNumA.length )
{
a[i] = bigNumA[bigNumA.length -1 - i] -'0'
}
val b = IntArray(maxArrayLength)
for (iin 0 until bigNumB.length)
{
b[i] = bigNumB[bigNumB.length -1 - i] -'0'
}
var result = IntArray(maxArrayLength)
for (iin 0 until maxArrayLength){
var tem = result[i]
tem += a[i]
tem += b[i]
if (tem >=10)
{
tem -=10
result[i+1] =1
}
result[i] = tem
}
var sb = StringBuilder()
var findFirst =false
//再次逆序取出
for ( iin result.size -1 downTo 0)
{
if (!findFirst){
if (result[i] ==0){
continue
}
findFirst =true
}
sb.append(result[i])
}
println(sb.toString())
}
